Answer: 0.2885J/g°c
Explanation:
Loss of heat of metal = Gain of heat by the water
Therefore
-Qm = +Qw
Where
Q = mΔTCp
Q = heat
M= mass
ΔT = T.f - Ti
Ti= initial temperature
T.f= final temperature
Cp= Specific heat (m is metal, w is water)
-Qm = +Qw
-[m(T.f-Ti)Cpm] = m(T.f-Ti)Cpw
Therefore
-[83.2g(28.42-90.31)Cpm] = 41.82g(28.42-
19.93)4.194J/g°c
Cpm = 1485.54÷ 5149.25
Cpm = 0.2885J/g°c
Therefore , the specific heat of the metal is 0.2885J/g°c
Answer:
A
1. density and composition
2. seismic waves
3.mantle
B.
1. lithosphere
2. asthenosphere
3. mantle
4. outer core
5. inner core
C.
box a I think
Answer: <em>Running water, even at a trickle, helps prevent pipes from freezing. If you decide to use fuel-burning equipment, such as a kerosene heater, ensure it is vented to outside and kept clear of any vents. Keep garage doors closed if there are water supply lines in the garage.</em>
Explanation:
<u>Answer:</u> The volume when the pressure and temperature has changed is 1332.53 L
<u>Explanation:</u>
To calculate the volume when temperature and pressure has changed, we use the equation given by combined gas law. The equation follows:
where,
are the initial pressure, volume and temperature of the gas
are the final pressure, volume and temperature of the gas
We are given:
![P_1=0.950atm\\V_1=200L\\T_1=27^oC=[273+27]K=300K\\P_2=0.125atm\\V_2=?L\\T_2=-10^oC=[273-10]K=263K](https://tex.z-dn.net/?f=P_1%3D0.950atm%5C%5CV_1%3D200L%5C%5CT_1%3D27%5EoC%3D%5B273%2B27%5DK%3D300K%5C%5CP_2%3D0.125atm%5C%5CV_2%3D%3FL%5C%5CT_2%3D-10%5EoC%3D%5B273-10%5DK%3D263K)
Putting values in above equation, we get:
Hence, the volume when the pressure and temperature has changed is 1332.53 L
Answer:
either first or second if not them try d but I'm pretty sure a also I'm sorry if I getbyou this wrong I dearly apologize
